ASRJC_9758_2024_Prelim P2 Solutions

[Turn Over ANDERSON SERANGOON JUNIOR COLLEGE H2 MATHEMATICS JC2 Prelim Paper 2 (100 marks) 9758 16 Sept 2024 3 hours Additional Material(s): List of Formulae (MF26) CANDIDATE NAME CLASS / READ THESE INSTRUCTIONS FIRST Write your name and class in the boxes above. Please write clearly and use capital letters. Write in dark blue or black pen. HB pencil may be used for graphs and diagrams only. Do not use staples, paper clips, glue or correction fluid. Answer all the questions and write your answers in this booklet. Do not tear out any part of this booklet. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use an approved graphing calculator. Where unsupported answers from a graphing calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. All work must be handed in at the end of the examination. If you have used any additional paper, please insert them inside this booklet. The number of marks is given in brackets [ ] at the end of each question or part question. Question number Marks 1 2 3 4 5 6 7 8 9 10 11 Total This document consists of 13 printed pages and 3 blank pages.

2 Section A: Pure Mathematics [40 marks] 1 (a) 2 Let cot d cosecd x x = =−   ( ) 22 2 22 1 d 1 1 cosec d cot 1 cot x xx + =  − +     ( ) 2 22 2 2 1 cosec d cot cosec sin 1 dcos sin =  − =−        ( ) 2sin (cos ) d −=−    OR tan sec d=−    1 cos C=− +  2 1x Cx +=− + 2 (ai) ( ) 2 1 2 2 2 2 du du 98 49 18 8 4 9 2 (3 ) u u u uu − − + = ++   21 21 1 8 3ln(4 9 ) tan2 (2)(3) 2 1 4 3ln(4 9 ) tan2 3 2 uuC uuC − −  = + − +   = + − +  (ii) 3 1 Area required = dyx ( ) 1 20 9 2 1 d49 u uuu=  ++ ( ) 21 20 11 200 18 9=d 49 982 1 d + d 49 uu uu uuu u + + −= +   1 21 0 1 4 32 ln(4 9 ) tan (from part (i))2 3 2 uuu − = + + −  

3 [Turn Over 11 12 1 4 3 1 4= 2 ln13 tan 0 ln 4 tan (0)2 3 2 2 3 1 13 4 32 ln tan units2 4 3 2 −− −     + − − + −         = + −  (b) Given 22( 2) 4( 1) 4xy+ + − = 2 2 2( 2) 4[1 ( 1) ] 2 2 1 ( 1)x y x y+ = − −  + = − − The shaded region is bounded by the section of the ellipse where x ≤ −2. Hence 22 2 1 ( 1)xy=− − − − . Volume of solid formed ( ) 2 22 1 2 2 2 1 π4 (2 1) π d 16π π 2 2 1 ( 1) d xy yy = − − = − − − − −   39.6 units (From GC)= 3 (a) 6 2 4 5 3 2 7 4 3 AB →            =−=                and 5 6 1 6 5 1 10 7 3 BE → −           = − =                Consider 4 1 3 1 2 1